What is the rms velocity in m/s of the molecules in a sample of dinitrogen monoxide gas at 0°C? Round your answer to the nearest whole number, and do not include units.

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Answer:

The root-mean-square (rms) velocity of gas molecules can be calculated using the formula:

\[ v_{\text{rms}} = \sqrt{\frac{3kT}{m}} \]

Where:

- \( k \) is the Boltzmann constant (\(1.38 \times 10^{-23} \, \text{J/K}\)),

- \( T \) is the temperature in Kelvin,

- \( m \) is the molar mass of the gas in kilograms per mole.

For dinitrogen monoxide (N2O), the molar mass is approximately \(44.013 \, \text{g/mol}\).

Given that the temperature is \(0°C\), which is \(273.15 \, \text{K}\) in Kelvin, we can calculate:

\[ v_{\text{rms}} = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \, \text{J/K} \times 273.15 \, \text{K}}{0.044013 \, \text{kg/mol}}} \]

\[ v_{\text{rms}} \approx 513 \, \text{m/s} \]

Rounded to the nearest whole number, the rms velocity is approximately \(513 \, \text{m/s}\).

To find the root mean square (rms) velocity of molecules in a sample of dinitrogen monoxide gas at 0°C, we can use the formula:

v = √(3kT/m)

where:
v is the rms velocity,
k is the Boltzmann constant (1.38 x 10^-23 J/K),
T is the temperature in Kelvin,
m is the molar mass of dinitrogen monoxide (N2O).

First, let's convert the temperature from Celsius to Kelvin. To do that, we add 273 to the Celsius temperature:

0°C + 273 = 273K

The molar mass of dinitrogen monoxide (N2O) is approximately 44 g/mol.

Now we can plug these values into the formula:

v = √((3 * 1.38 x 10^-23 J/K * 273K) / 0.044 kg/mol)

After calculating this, the rms velocity of the molecules in the sample of dinitrogen monoxide gas at 0°C is approximately 487 m/s.

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